If three consecutive vertices of a parallelogram are $A(4,3,5), B(0,6,0)$, $C(-8,1,4)$ and $D$ is the fourth…

If three consecutive vertices of a parallelogram are $A(4,3,5), B(0,6,0)$, $C(-8,1,4)$ and $D$ is the fourth vertex, then the angle between $\mathbf{A C}$ and $\mathbf{B D}$ is
  1. $\cos ^{-1}\left(\frac{65}{\sqrt{149} \sqrt{161}}\right)$
  2. $\cos ^{-1}\left(\frac{55}{\sqrt{149} \sqrt{161}}\right)$
  3. $\cos ^{-1}\left(\frac{73}{\sqrt{149} \sqrt{161}}\right)$
  4. $\cos ^{-1}\left(\frac{15}{\sqrt{149} \sqrt{161}}\right)$

Solution

Given, $A(4,3,5), B(0,6,0), C(-8,1,4)$ and $D$ are the vertices of a parallelogram Let $D$ be the point $(x, y, z)$. $\therefore$ Diagonals of parallelogram bisect each other.
$\Rightarrow$ mid-point of $A C=$ mid point of $B D$ $\begin{aligned} & \left(\frac{4+(-8)}{2}, \frac{3+1}{2}, \frac{5+4}{2}\right)=\left(\frac{x+0}{2}, \frac{y+6}{2}, \frac{z+0}{2}\right) \\ & \Rightarrow \quad\left(-2,2, \frac{9}{2}\right)=\left(\frac{x}{2}, \frac{y+6}{2}, \frac{z}{2}\right) \\ & \Rightarrow \quad \frac{x}{2}=-2, \quad \frac{y+6}{2}=2, \quad \frac{z}{2}=\frac{9}{2} \\ & x=-4, \quad y=-2, \quad z=9\end{aligned}$ $\therefore D$ is $(-4,-2,9)$. Now, $\mathbf{A C}=\mathrm{PV}$ of $C-\mathrm{PV}$ of $\mathrm{A}$ $ \begin{aligned} & =(-8 \hat{\mathbf{i}}+\hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \\ & =-8 \hat{\mathbf{i}}+\hat{\mathbf{j}}+4 \hat{\mathbf{k}}-4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\ & =-12 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ \mathbf{B D} & =\mathrm{PV} \text { of } D-\mathrm{PV} \text { of } B \\ & =(-4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+9 \hat{\mathbf{k}})-(0 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+0 \hat{\mathbf{k}}) \\ & =-4 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+9 \hat{\mathbf{k}} \end{aligned} $ Let $\theta$ be the angle between AC and BD. $\begin{aligned} & \text { Then, } \cos \theta=\left|\frac{\mathbf{A C} \cdot \mathbf{B D}}{|\mathbf{A C}||\mathbf{B D}|}\right| \\ & =\left|\frac{(-12 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(-4 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+9 \hat{\mathbf{k}})}{\sqrt{(12)^2+(2)^2+(1)^2} \sqrt{(4)^2+(8)^2+(9)^2}}\right| \\ & \Rightarrow \cos \theta=\left|\frac{48+16-9}{\sqrt{161} \sqrt{149}}\right| \\ & \Rightarrow \quad \theta=\cos ^{-1}\left(\frac{55}{\sqrt{149} \sqrt{161}}\right)\end{aligned}$

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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